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Can we say that a set $E\subset \mathbb{R}^n$ is convex iff $E\cap L$ is contractible for every real affine line $L$ in $\mathbb{R}^n$?

I know that $E\subset \mathbb{R}^n$ is convex iff $E\cap L$ is convex for every real affine line $L$ in $\mathbb{R}^n$. And a convex set is contractible.

Also if we consider $E\subset \mathbb{R}^n\subset \mathbb{C}^n$ , then is $E$ convex if and only if $E\cap L$ is contractible for every complex affine line $L$ in $\mathbb{C}^n$?

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Strictly speaking this is not true since $\emptyset$ is not contractible. So let us interpret the condition in the sense that for every real affine line $L$ in $\mathbb{R}^n$, the set $E\cap L$ is either empty or contractible.

You know that $E$ is convex iff $E\cap L$ is convex for every real affine line $L$ in $\mathbb{R}^n$. Thus your first question is equivalent to showing that a subset $A \subset \mathbb R$ is convex iff it is contractible.

Clearly nonempty convex sets are contractible. Now assume that $A$ is not convex. This means that there exist $x,y \in A$ such that $z_t = tx + (1-t)y \notin A$ for some $t \in [0,1]$. W.l.o.g. we may assume that $x \le y$. It is impossible that $x = y$ because then $z_t = x$ for all $t \in [0,1]$. Thus $x < y$. Hence $A$ is not connected because $A \cap (-\infty, z_t)$ and $A \cap (z_t,\infty)$ form a partition of $A$ in disjoint nonempty open sets. But a non-connected set is not contractible.

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